Question

A person standing on a 20 m high building throws ball up in the air with a velocity of 14 m per 2nd period he goes to a maximum high and falls to the ground. How long does it take for the ball to reach maximum height? What is the balls maxim height? What is the ball's final velocity before hitting the ground

Answer 1

(a)

$$t=\frac{v_i}{g}=\frac{14}{9.8}=1.4\;\rm{s}$$

(b) The maximum height above the building

$$h_{max}=\frac{v_i^2}{2g}=\frac{14^2}{2\times 9.8}=10\;\rm{m}$$

The maximum height above the ground

$$h=10\:\rm{m}+20\:\rm{m}=30\;\rm{m}$$

(c) The final velocity

$$v_f=\sqrt{2gh}=\sqrt{2\times 9.8\times 30}=54\;\rm{m/s}$$